Answer:
3.03 * 10 ⁻⁴ m/s
Explanation:
see attached picture for explanation.
Answer:
500 N
Explanation:
Given;
Mass of the car, M = 1000 kg
initial speed of the car, u = 0 m/s
Final speed of the car, v = 60 m/s
Time, t = 1 min = 60 s
Now,
Force, F is given as:
F = Ma
where,
a is the acceleration
From the Newton's equation of motion, we have
v = u + at
on substituting the values, we get
60 = 0 + a × 60
or
a = 1 m/s²
Thus,
Force = 1000 × 1 = 1000 N
now,
this force will be equal to the friction force provided by the rear wheels
let the friction force on a single rear wheel be 'f'
thus,
2f = 1000 N
or
f = 500 N
Answer: 133.88 MPa approximately 134 MPa
Explanation:
Given
Plane strains fracture toughness, k = 26 MPa
Stress at which fracture occurs, σ = 112 MPa
Maximum internal crack length, l = 8.6 mm = 8.6*10^-3 m
Critical internal crack length, l' = 6 mm = 6*10^-3 m
We know that
σ = K/(Y.√πa), where
112 MPa = 26 MPa / Y.√[3.142 * 8.6*10^-3)/2]
112 MPa = 26 MPa / Y.√(3.142 * 0.043)
112 = 26 / Y.√1.35*10^-2
112 = 26 / Y * 0.116
Y = 26 / 112 * 0.116
Y = 26 / 13
Y = 2
σ = K/(Y.√πa), using l'instead of l and, using Y as 2
σ = 26 / 2 * [√3.142 * (6*10^-3/2)]
σ = 26 / 2 * √(3.142 *3*10^-3)
σ = 26 / 2 * √0.009426
σ = 26 / 2 * 0.0971
σ = 26 / 0.1942
σ = 133.88 MPa